专题06 机械能守恒定律 能量守恒定律
目 录
01机械能转化及守恒的判断 2
考向一：机械能转化及守恒的判断 2
02 机械能守恒定律的应用 7
考向一：机械能的图像类问题 7
考向二：单个物体的机械能守恒 15
考向三：多个物体(系统)的机械能守恒 18
03能量转化及守恒定律的综合应用 20
考向一：能量转化及守恒定律的综合应用 20
考向四：涉及弹簧的能量问题 22
考向二：涉及板块、传送带的能量问题 33
01机械能转化及守恒的判断
考向一：机械能转化及守恒的判断
1．（2023·浙江·高考真题）一位游客正在体验蹦极，绑上蹦极专用的橡皮绳后从跳台纵身而下。游客从跳台下落直到最低点过程中（ ）
A．弹性势能减小 B．重力势能减小
C．机械能保持不变 D．绳一绷紧动能就开始减小
3．（2022·浙江·高考真题）神舟十三号飞船采用“快速返回技术”，在近地轨道上，返回舱脱离天和核心舱，在圆轨道环绕并择机返回地面。则（ ）
A．天和核心舱所处的圆轨道距地面高度越高，环绕速度越大
B．返回舱中的宇航员处于失重状态，不受地球的引力
C．质量不同的返回舱与天和核心舱可以在同一轨道运行
D．返回舱穿越大气层返回地面过程中，机械能守恒
5．（2021·浙江·高考真题）如图所示，同学们坐在相同的轮胎上，从倾角相同的平直雪道先后由同高度静止滑下，各轮胎与雪道间的动摩擦因数均相同，不计空气阻力。雪道上的同学们（ ）
A．沿雪道做匀速直线运动 B．下滑过程中机械能均守恒
C．前后间的距离随时间不断增大 D．所受重力沿雪道向下的分力相同
7．（2024·广东江门·统考模拟预测）蹦极是一项非常刺激的体育项目，而“反向蹦极”是一项比蹦极更加刺激的运动。如图所示，弹性轻绳上端固定在P点，下端固定在玩家的身上并向下拉长与固定在地面上的安全扣环相连。打开安全扣环后，玩家从c点开始由静止释放，像“火箭发射一般竖直向上运动”，上升到b点时速度最大，轻绳在a点为原长位置，玩家到达的最高点在P与a之间，不计空气阻力，下列说法正确的是（ ）
A．人在c到a运动过程中都做加速运动
B．人在c到b运动过程中做加速度减小的加速运动
C．在整个运动过程中，玩家的机械能守恒
D．人在b到最高点的运动过程中，玩家处于超重状态
8．（2024·广东惠州·统考二模）2023年10月3日，在杭州亚运会蹦床项目女子决赛中，中国选手朱雪莹夺冠，图为朱雪莹在东京奥运会上决赛时腾空后下落的照片，朱雪莹从刚接触床面到运动至最低点的过程中，下列说法正确的是（ ）
A．朱雪莹刚接触床面时速度最大
B．朱雪莹在该过程中始终处于失重状态
C．在该过程中朱雪莹的动能减少量等于弹性势能的增加量
D．在该过程中蹦床对朱雪莹一直做负功
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晏全萍2986天体椭圆运动可以看成是简谐振动么 -
殷庆丁13114323165 ______ 很明显不行,卫星如果绕某一中心做椭圆运动,其中中心是椭圆的焦点,关于长轴对称,但明显不是简写运动,振幅不等啊
晏全萍2986平行运动是怎么回事?
殷庆丁13114323165 ______ 一种是汽车上的货物和车厢的运动,叫 做平动.货物无论放在车厢上层还是下层, 车厢前面或者后面,都和车厢一起沿汽车前进的方向上作平行移动.它们不仅通过的距 离相等,而且运动快慢和方向也都相同,如在一秒钟内都向前平行移动10米的距离.这 种形式的运动叫平动.木工创木板时创子的运动,错工里工件时健刀的运动,把抽屉从桌 子里拉出来的运动等,都属于平行运动.
晏全萍2986进行能量转换的主要场所是哪?进行能量转换的主要场所是哪里
殷庆丁13114323165 ______ 线粒体进行能量转换.
晏全萍2986一切物体都具有波动性吗?是必须是运动的物体才有波动性的吗? -
殷庆丁13114323165 ______ 所有的物体都有波动性,人也不例外
晏全萍2986在水中传播的时候哪一种波在传播过程中能量损失最少? -
殷庆丁13114323165 ______ 声波
晏全萍2986内能少的物体也能将能量传给内能多的物体,这句话是正确的还是错误的,再举一个例子证明你的观点 -
殷庆丁13114323165 ______ 这句话正确. 例如1g 100℃的水,内能显然小于1000kg 99℃的水的内能,但这里1g水的温度较高,可以将内能传给 1000k...
晏全萍2986动量水平传送带维持恒定的速率v=1.0m/s不变,位于起始端的料
殷庆丁13114323165 ______ 系统不断在将速度为0的煤加速,结果是让他们拥有了v = 1.0 m/s的速度. 2.4*10^3KG/MIN=40KG/S 在时间T中,有40*T KG的煤被加速 需要的传送力是 F*T=M*V F=40*T*1/T=40N 所以做的功是 1/2*40*T*1^2=20T 功率是20T/T=20W 第一问应该是40N 第二问我想错了.除了给煤的动能外,还有摩擦造成的热能损耗.因此结果错误. 实际功率应该是 P=F*V =40N*1M/S =40W
